Many practical problems involve finding a solution to achieve a maximum or minimum value such as maximising profit, minimising costs, maximising volume of a box or minimising distance travelled. The process to find the best solution is called optimisation. There are many optimisation techniques for different problems and in many cases where we can describe the outcome with a function we can use calculus to solve for the optimal solution.
In a previous lesson, we have seen how to identify and classify key features in a function using calculus. We can now apply these to problems in context and in particular, identifying maximum and minimum values of a function.
Problem solving steps for optimisation applications
- Where appropriate, draw a clear diagram to illustrate the problem. Label the diagram and define any variables. Note if the variable has any restrictions including implied restrictions - such as the physical dimensions of an object need to be positive.
- Write an expression for the quantity to be optimised using known relationships. Express this quantity as a function in terms of a single variable - this may require some algebraic manipulation such as substitution.
- Find the derivative of your function and solve for the value(s) which make the derivative equal zero.
- Determine the nature of any stationary points found - local maximum, local minimum or stationary point of inflection. (This can be done by checking the value of the derivative either side of any stationary points or by using the second derivative to check concavity. It is a good idea to confirm by graphing the function using technology.)
- Identify the optimal solution, remembering to check end points if the domain is restricted and reflect if the solution is practical.
Worked example
A piece of wire, $320$320 $cm$cm long, is used to make the twelve edges of a rectangular box in which the length is three times the width.
a) Find the maximum volume of the box that can be formed using the wire and the dimensions of this optimal box.
Step 1. Draw and label a diagram of the situation.
Let $x$x be the width of the box and $h$h be the height of the box. The length of the box is given by $3x$3x. Since we will require the volume in terms of one variable, let's rewrite $h$h in terms of $x$x.
| $\text{Total length of wire}$Total length of wire | $=$= | $4\left(3x+x+h\right)$4(3x+x+h) |
| $320$320 | $=$= | $4\left(4x+h\right)$4(4x+h) |
| $4x+h$4x+h | $=$= | $80$80 |
| $\therefore h$∴h | $=$= | $80-4x$80−4x |
Note: Since length, width and height need to be positive we have the implied restriction: $0\le x\le20$0≤x≤20
Step 2. We want to maximise the volume, so we want to express the volume in terms of $x$x.
| $\text{Volume}$Volume | $=$= | $\text{length}\times\text{width}\times\text{height}$length×width×height | using known formulas |
| $V\left(x\right)$V(x) | $=$= | $3x\times x\times\left(80-4x\right)$3x×x×(80−4x) | |
| $=$= | $240x^2-12x^3$240x2−12x3 $m^2$m2 | expanding as this will be required for differentiation |
Step 3. Find the derivative and solve $V'\left(x\right)=0$V′(x)=0:
The derivative is $V'\left(x\right)=480x-36x^2$V′(x)=480x−36x2. When $V'\left(x\right)=0$V′(x)=0 we have:
| $480x-36x^2$480x−36x2 | $=$= | $0$0 |
| $4x\left(120-9x\right)$4x(120−9x) | $=$= | $0$0 |
Therefore, by the null factor law, either:
| $4x$4x | $=$= | $0$0 | or | $120-9x$120−9x | $=$= | $0$0 |
| $\therefore x$∴x | $=$= | $0$0 | $-9x$−9x | $=$= | $-120$−120 | |
| $\therefore x$∴x | $=$= | $13\frac{1}{3}$1313 |
When$x=0$x=0, $V\left(0\right)=0$V(0)=0 $cm^3$cm3 and when $x=13\frac{1}{3}$x=1313, $V\left(13\frac{1}{3}\right)=14222\frac{2}{9}$V(1313)=1422229 $cm^3$cm3. The volume of $0$0 $m^3$m3 is clearly not a maximum and in fact we do not have a box when the width and length are $0$0 $cm$cm. So let's simply check the nature of the stationary point at $\left(13\frac{1}{3},14222\frac{2}{9}\right)$(1313,1422229).
Step 4. To confirm this is a maximum we can test the derivative either side of the stationary point:
| $x$x | $10$10 | $13\frac{1}{3}$1313 | $15$15 |
|---|---|---|---|
| $f'\left(x\right)$f′(x) | $1200$1200 | $0$0 | $-900$−900 |
| Sign | $+$+ | $0$0 | $-$− |
| Shape |  |
 |
 |
Alternatively, calculate the second derivative and test it is negative (concave down) at this point or use information about the form of the graph (cubic with a negative leading coefficient) to determine the nature of the stationary point.
Step 5. Hence, we have a maximum volume of $14222.\overline{2}$14222.2 $m^3$m3 when the dimensions of the box are $13\frac{1}{3}$1313 $cm$cm by $40$40 $cm$cm by $26\frac{2}{3}$2623 $cm$cm. This can also be confirmed by graphing the function on the appropriate domain.
b) If the height of the box can be at most $20$20 $cm$cm, what is the maximum volume that can be formed using the wire and the dimensions of this optimal box.
If the height is restricted we have the following inequality:
| $0$0 | $\le$≤ | $80-4x$80−4x | $\le$≤ | $20$20 |
| $-80$−80 | $\le$≤ | $-4x$−4x | $\le$≤ | $-60$−60 |
| $20$20 | $\ge$≥ | $x$x | $\ge$≥ | $15$15 |
| $\therefore15$∴15 | $\le$≤ | $x$x | $\le$≤ | $20$20 |
Sketching the graph on the restricted domain we can see the maximum occurs at the end point, $x=15$x=15.
Hence, with the height restriction the maximum volume of the box is $13500$13500 $cm^3$cm3, when the dimensions are $15$15 $cm$cm by $45$45 $cm$cm by $20$20 $cm$cm.
Practice questions
Question 1
If $A=xy$A=xy and $20x+5y=160$20x+5y=160:
Find an expression for $A$A in terms of $x$x only. Leave your answer in expanded form.
Find the possible value of $x$x that maximises $A$A.
Find an expression for the second derivative, $A''$A′′.
Select the option that best describes the nature of the curve at $x=4$x=4.
Since the value of $A''$A′′ at $x=4$x=4 is $8$8, the value of $A$A is minimised when $x=4$x=4.
ASince the value of $A''$A′′ at $x=4$x=4 is $-8$−8, the value of $A$A is maximised when $x=4$x=4.
BSince the value of $A''$A′′ at $x=4$x=4 is $8$8, the value of $A$A is maximised when $x=4$x=4.
CSince the value of $A''$A′′ at $x=4$x=4 is $-8$−8, the value of $A$A is minimised when $x=4$x=4.
DState the maximum value of $A$A.
Hence calculate the value of $y$y that maximises $A$A.
Question 2
The concentration, $x$x, of a certain medication in the bloodstream of a patient, $t$t hours after taking a dose, is given by $x=8te^{-4t}$x=8te−4t.
What is the concentration in the bloodstream when the patient initially takes the dose?
Solve for the time $t$t at which the patient has the greatest concentration of medication in his bloodstream.
Determine the maximum concentration of the medication. Round your answer to three decimal places.
According to the model, does the concentration ever reach $0$0 again?
Yes
ANo
B
Question 3
In the diagram below, the function $y=\frac{b}{x^2+a}$y=bx2+a has a maximum turning point at $\left(0,4\right)$(0,4) and passes through $\left(3,1\right)$(3,1).
A rectangle $BCDE$BCDE is inscribed within the curve as shown. $OY$OY is the axis of symmetry.
Using the fact that the function passes through $\left(0,4\right)$(0,4) find an expression for $b$b in terms of $a$a.
Using the results of part (a) and the fact that the function passes through $\left(3,1\right)$(3,1), find the value of $a$a.
Using the results of part (a) and (b), find the value of $b$b.
If $C$C has coordinates $\left(p,0\right)$(p,0), find the $y$y coordinate of $B$B in terms of $p$p.
Find the area $A(p)$A(p) of $BCDE$BCDE in terms of $p$p.
Find the $p$p values for all turning points of $A\left(p\right)$A(p). Write each solution on the same line, separated by commas.
You may use the substitutions $u=24p$u=24p and $v=p^2+3$v=p2+3.
By filling in the blanks, show that $p=\sqrt{3}$p=√3 results in the maximum area of $BCDE$BCDE, and find that maximum area. Round any necessary answers to two decimal places.
$p$p $1$1 $\sqrt{3}$√3 $2$2 $A'\left(p\right)$A′(p) $\editable{}$ $\editable{}$ $\editable{}$ Maximum area: $A\left(\sqrt{3}\right)=\editable{}$A(√3)= units2